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Cell[CellGroupData[{
Cell["Arc Length & Curvature", "Title",
  TextAlignment->Center,
  FontColor->RGBColor[0, 0, 1]],

Cell[TextData[StyleBox["Arc Length",
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  FontColor->RGBColor[0, 0, 1]]], "Subtitle"],

Cell[TextData[{
  "Let ",
  StyleBox["r",
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  "(t) be a vector valued function whose graph is a space curve\n\t",
  StyleBox["r",
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  "(t) = x(t) ",
  StyleBox["i",
    FontWeight->"Bold"],
  " + y(t) ",
  StyleBox["j",
    FontWeight->"Bold"],
  " + z(t) ",
  StyleBox["k\n",
    FontWeight->"Bold"],
  "Then the length of the space curve over the interval a \[LessEqual] t \
\[LessEqual] b (assuming the curved is traced only once from t = a to t = b) \
is given by:\n\tL = ",
  Cell[BoxData[
      \(TraditionalForm
      \`\[Integral]\_a\%b
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          \[DifferentialD]t\)]],
  "\n\t\n\tor\n\t\n\tL = ",
  Cell[BoxData[
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                  FontWeight->"Bold"]}], \((t)\)}]}], "|", 
          \(\[DifferentialD]t\)}], TraditionalForm]]],
  "\nUsually, it is very difficult to find this definite integral, unless the \
space curve is a simple curve (like a line or a circular helix).  ",
  StyleBox["Mathematica",
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  " helps greatly in performing the integtation for arc length of more \
complex space curves."
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Cell[TextData[{
  "Example 1\nFind the arc length of the circular helix with vector valued \
equation ",
  StyleBox["r",
    FontWeight->"Bold"],
  "(t) = 3cos t ",
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    FontWeight->"Bold"],
  " + 5t ",
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  StyleBox[" k",
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  " from t = 0 to t = 2."
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  "\nIn Example 1, the integrand was actually very simple.  Using the trig. \
identity ",
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Cell[TextData[{
  "Example 2\nFind the arc length of the elliptical helix with vector valued \
equation ",
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  "(t) = 3cos t ",
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    FontWeight->"Bold"],
  " + 5t ",
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  "from t = 0 to t = 2.  "
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Mathematica is finding the exact value of the definite integral \
using an elliptic integral with which you are unfamiliar, but we can still \
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\>", "Subtitle",
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  "(t) by its length we get the Unit Tangent vector, ",
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  "Example 3\nFind the unit Tangent vector to the circular helix with vector \
valued equation ",
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Cell[TextData[{
  "The curvature of a curve at a given point is a measure of how quickly the \
curve changes direction at that point. For instance, a straight line has \
curvature of zero, and a circle of radius a has a constant curvature of 1/a.  \
 The symbol for curvature is \[Kappa]  and it is defined to be the magnitude \
of the rate of change of the Unit Tangent vector with respect to the arc \
length.   \n\n\t\[Kappa] = |",
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Cell[TextData[{
  "This definition of curvature has the advantage that the curvature is \
independent of the parameter t, but it requires that we parametrize the \
vector valued function with respect to the arc length - something we will not \
cover in this class.  So we need other ways to express the curvature.  Below \
you will find two more useful expressions for curvature: \n\n\t\[Kappa] (t) = \
",
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                StyleBox[\((t)\),
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r(t) and its derivatives \n\t\n\t\[Kappa] (t) = ",
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Cell[TextData[{
  "Example 4\nFind the curvature of the circular helix with vector valued \
equation ",
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  StyleBox["Assignment for Lab 4",
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